Question

670 nm light shines through two slits that are separated by 0.3x10-3 m, and forms a striped pattern on a viewing screen. The fourth bright line on the viewing screen is located 0.05 m from the center of the pattern. What is the distance from the slits to the viewing screen in meters?

Answer 1

This looks like a physics problem involving the double-slit experiment. According to the web sources¹², the formula for finding the distance from the slits to the viewing screen (denoted by L) is:

L = (xd)/(n*lambda)

where x is the distance from the center of the pattern to the nth bright line, d is the separation between the slits, n is the order number of the bright line, and $\lambda$ is the wavelength of light.

Plugging in the given values, we get:

L = (0.05m)(0.3 x 10 ^-3 m)/(4)(670 x 10^-9 m)

Simplifying, we get:

L = 5.597

So, the distance from the slits to the viewing screen is approximately 5.6 meters.

Answer 2

We can use the equation for the location of the bright fringes in a double-slit experiment:

y = (mλD) / d

where y is the distance from the central maximum to the m-th bright fringe, λ is the wavelength of the light, d is the distance between the two slits, and D is the distance from the slits to the viewing screen.

We are given λ = 670 nm, d = 0.3x10^-3 m, m = 4, and y = 0.05 m (the distance from the center to the fourth bright fringe). Therefore, we can solve for D:

D = (y * d) / (m * λ) = (0.05 m * 0.3x10^-3 m) / (4 * 670x10^-9 m) ≈ 1.12 m

the distance from the slits to the viewing screen is approximately 1.12 meters.